Other factors of the equation $2x^4 + 4x^3\sin A\sin B -\cdots$

Question

If $x^2+x\cos(A + B) + 1$ is a factor of the expression, $$2x^4 + 4x^3\sin A\sin B -x^2(\cos 2A +\cos 2B) + 4x\cos A\cos B -2$$ Then we have to find the other factor (s).

I am not getting start

Can anybody provide me a hint .

Answer 1

Prerequisites: Recall the identities: $$2\sin A\sin B = \cos (A-B) -\cos (A+B) \tag {1} $$ $$2\cos A \cos B = \cos (A+B) + \cos (A-B) \tag {2} $$

Using these, our expression simplifies to $$2x^4 +2x^3\cos (A-B) -2x^3\cos (A+B) -2x^2\cos (A+B)\cos (A-B) +2x\cos(A+B)+ 2x\cos (A-B)-2$$

Hope you can take it from here.

Answer 2

Then the other factor shall be a quadratic polynomial that can be generally formulated as $$ C\,x^{\,2} + D\,x + E $$ So you must have $$ \begin{gathered} \left( {C\,x^{\,2} + D\, x + E} \right)\left( {\,x^{\,2} + \,\cos (A + B)\,x + 1} \right) = \hfill \\ = 2x^{\,4} + \left( {4\sin A\sin B} \right)\,x^{\,3} - \left( {\cos 2A + \cos 2B} \right)x^{\,2} + \left( {4\cos A\cos B} \right)\,x - 2 \hfill \\ \end{gathered} $$ and equating the coefficients of the powers of x $$ \left\{ \begin{gathered} C = 2 \hfill \\ C\cos (A + B) + D = 4\sin A\sin B \hfill \\ C + D\cos \left( {A + B} \right) + E = - \left( {\cos 2A + \cos 2B} \right) \hfill \\ D + E\cos (A + B) = 4\cos A\cos B \hfill \\ E = - 2 \hfill \\ \end{gathered} \right. $$ Now, adding 2nd row+4th row gives: $$ D = 2\left( {\cos A\cos B + \sin A\sin B} \right) = 2\cos (A - B) $$ However, this does not satisfy any of the three middle row !
That means that $\left( {\,x^{\,2} + \,\cos (A + B)\,x + 1} \right)$ is not a factor of the quartic binomial, for any generic $A$ and $B$. Substituting $D = 2\cos (A - B)$ back into the system, you get that it is totally satisfied iff $$ \left\{ \begin{gathered} \cos (A + B) = 0 \hfill \\ \cos (2A) + \cos (2B) = 0 \hfill \\ \end{gathered} \right. $$ i.e., if $$ \left\{ \begin{gathered} A + B = \pi /2 + 2k\pi \hfill \\ 2A = 2B + \pi + 2j\pi \hfill \\ \end{gathered} \right.\quad \left\{ \begin{gathered} A + B = \pi /2 + 2k\pi \hfill \\ A - B = \pi /2 + j\pi \hfill \\ \end{gathered} \right.\quad \left\{ \begin{gathered} A = \left( {2k + j + 1} \right)\pi /2 \hfill \\ B = \left( {2k - j} \right)\pi /2 \hfill \\ \end{gathered} \right. $$